Simplifying if statement

[Simbaku]

So I am looking back on a tic tac toe project I worked on for a previous class, and found something that bugs me now as it did when I worked on it. It is this large clump of code:

	if (arrStorage[0][0] == 'X' && arrStorage[0][1] == 'X' && arrStorage[0][2] == 'X' || //X Across first row
		arrStorage[1][0] == 'X' && arrStorage[1][1] == 'X' && arrStorage[1][2] == 'X' || //X Across Second Row
		arrStorage[2][0] == 'X' && arrStorage[2][1] == 'X' && arrStorage[2][2] == 'X' || //X Across Third Row
		arrStorage[0][0] == 'X' && arrStorage[1][0] == 'X' && arrStorage[2][0] == 'X' || //X Down  First  Row
		arrStorage[0][1] == 'X' && arrStorage[1][1] == 'X' && arrStorage[2][1] == 'X' || //X Down  Second Row
		arrStorage[0][2] == 'X' && arrStorage[1][2] == 'X' && arrStorage[2][2] == 'X' || //X Down  Third  Row
		arrStorage[0][0] == 'X' && arrStorage[1][1] == 'X' && arrStorage[2][2] == 'X' || //X Diagonal Top left
		arrStorage[0][2] == 'X' && arrStorage[1][1] == 'X' && arrStorage[2][0] == 'X' || //X Diagonal Top Right

		arrStorage[0][0] == 'O' && arrStorage[0][1] == 'O' && arrStorage[0][2] == 'O' || //O Across first row
		arrStorage[1][0] == 'O' && arrStorage[1][1] == 'O' && arrStorage[1][2] == 'O' || //O Across Second Row
		arrStorage[2][0] == 'O' && arrStorage[2][1] == 'O' && arrStorage[2][2] == 'O' || //O Across Third Row
		arrStorage[0][0] == 'O' && arrStorage[1][0] == 'O' && arrStorage[2][0] == 'O' || //O Down  First  Row
		arrStorage[0][1] == 'O' && arrStorage[1][1] == 'O' && arrStorage[2][1] == 'O' || //O Down  Second Row
		arrStorage[0][2] == 'O' && arrStorage[1][2] == 'O' && arrStorage[2][2] == 'O' || //O Down  Third  Row
		arrStorage[0][0] == 'O' && arrStorage[1][1] == 'O' && arrStorage[2][2] == 'O' || //O Diagonal Top left
		arrStorage[0][2] == 'O' && arrStorage[1][1] == 'O' && arrStorage[2][0] == 'O')	//O Diagonal Top Right
	{

I have tried experimenting with ways to significantly shorten this without getting overtly complicated but I am having no luck. I would love some advice as to how I could approach simplifying this.

-The Array is a 2 dimensional [3][3]

[ne555]

for(int K=0; K<n; ++K)
   array[row][K] //traverse row
   array[K][col] //traverse column
   array[K][K] //traverse one diagonal
   array[K][n-K-1] //traverse the other diagonal 

in each traverse you check if all the symbols are equal

[jonnin]

What I would do is..

when the user takes a turn,
the square that was modified on that turn,
check that square's row, column, and if applicable, diagonals for a win.

That would simplify the logic greatly. First, you don't check things not affected by the turn, and second, you only check the letters for the turn-taker, cutting work in half. (that is, if X just moved, you don't need to know if O just won the game). Count the # of turns taken. If turn == 9 and didnt just win, is drawn.